Can someone help me please
2 answers:
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Answer:
Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?
Explanation:
So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s
so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull
Answer:
your friend did a work of 1717.34 J
Explanation:
The sum of the work made by your friend and you is equal the change in the kinetic energy, so:
Where is the work that you did,
is the work that your friend did, m is the mass of the satellite and
and
are the final and initial speeds
Additionally the work that you did is equal to:
Where F and s are vectors. So, the vector s that said the displacement is calculated as:
s = <2.3, 2.6, -11.8> - <3.3, -1.6, 2.0>
s = <-1 , 4.2 , -13.8>
Therefore, is:
Then, replacing the values on initial equation and solving for , we get:
Finally, your friend did a work of 1717.34 J